JT effects to calculate lowest temperature 2

[rafcu]

Dear Sirs,

I wonder if the JT effect that occurs in the orifice plate to depressurize natural gas from a pressure of 70 bar a is enough to calculate the gas lowest temperature at downstream orifice flow.

My concern come from the fact that the JT effect does not consider the kinematics energy required for increase the velocity till the choked velocity; the phenomena produce a not complete isenthalpic expansion, then, I conclude, that the temperature will be lower than calculated considering just JT effect.

Can anybody help me please?

 

[MortenA]

On the other hand theres the heat flux from the pipe into the gas?

Maybe this is a bit over the top?

If you are concerned and want a conservative measure why dont you just calculate the adiabatic temperature drop and use that for the inlet value to the orifice? Since you dont have choked flow upstream your orifice i would believe your pressuredrop is not that big?


Best regards

Morten

 

[25362]

Visiting thread391-78680 you may find some help.

 

[israelkk]

I assume that when you say natural gas you mean Methane. JT effect happens at constant enthalpy. Therefore, if you check the enthalpy at room temperature (300 Kelvin) and 70 bar it has to be equal to the enthalpy at 1 bar after the orifice. Therefore, the temperature of the Methane at 1 bar and the same enthalpy is ~268.8 Kelvin (-4.36 Celsius)

 

[sailoday28]

Are you considering temperature at the "throat-vena contractor" or somewhat downstream of the orifice?
At the throat, the KE change is substantial.
If flow and downstream pressure are known, a reasonable approx is to use adiabatic flow to determine the temperature.
The analysis will be more complicated if heat transfer is to be accounted for as Morten A suggests.

 

[25362]

Although rafcu refers to discharge through an orifice, the quotation I'm bringing herebelow from a short section of a Chem Eng article dated October 25, 1976 by Les Driskell titled Coping with high-pressure letdown, may be of help:

When a gas flows through a throttling valve, the overall thermodynamic process is isenthalpic, but from inlet to the vena contracta- the point of minimum cross-sectional area of the stream- the process is nearly isentropic. For this reason , the temperature at the vena contracta can be appreciably lower than that at the valve outlet, depending on the thermodynamic properties of the fluid. Ethylene, and even superheated steam, can condense precipitously within the valve body to create droplets of high velocity liquid. Valve construction materials must be selected to withstand the erosive scrubbing action of such streams, and measures must be taken to abate the added noise that may result.
Solids in the form of ice or hydrates may be induced during pressure letdown by auto-refrigeration if water is present...

 

[rafcu]

Dear Sirs,

Thank you for your kind help;

25362, I found the theoretical answers in your quote, please, could you send me the complete article?

Now, my concern is how calculate lowest temperature in the vena-contracta

Regards

 

[25362]

Rafcu, estimate the pressure at the vena contracta and move isentropically from the high pressure, easily done for a single component.

Provide a fax number and I'll try to send the article to you (6 pages). Disregard notes and marks.

 

[sailoday28]

25362 (Chemical) If the process is adiabatic, steady state,and has negligible change in elevation, then the process is approximated as isenthalpic only if there is neglibible change in upstream and downstream velocities.

Generally, it can be shown that the change in elevation has negligible effect. However,if change in upstream and downstream velocities is significant the isenthalpic assumption is not valid.

 

[sailoday28]

For adiabatic steady flow of a perfect gas, constant specific heats
Ao^2=A^2 + (gamma-1)/2 *U^2 (1)
where A=acoustic speed
U=local velocity
gamma=ration of specific heats, Cp/Cv

subscript o, stagnation conditions

at choked conditions (vena contracta) A=U, Mach=1

(1)yields (A/Ao)^2 = T/To= 2/(gamm+1)
where T is absolute temperature.

For gamma =1.4 T/To =1/1.2

If upstream stag temp is about 80F,
then T approx =450R, -10F

 

[25362]

May be some help can be obtained from

http://www.grc.nasa.gov/WWW/K-12/airplane/isentrop.html

and links therein

 

[rafcu]

Thanks for your support,

25362,

My fax number: +33 2 40 91 06 19 attentions to: Process Engineer

Exactly my problem is that my customer wants estimate the lowest temperature downstream and upstream orifice flow, for depressurizing a natural gas line at 70 bar a till 7 bar a at the worst temperature condition: The ESD start at -3°C,

Using Hysys simulation, depressurization utility, I obtain -60°C upstream (isentropic depressurization efficiency 90%) during the peak flow.

The problem is that Hysys does not consider the kinetic energy to arrive at sound velocity in orifice flow.

Some people recommended me use OLGA, but this software is not adapted to piping, is better for pipelines

The CS pipe was already damaged during start up. I will suggest use LTCS 3 meters upstream the OF, but the design temperature for this material is -49°C, I hope that the material allowances allows lower temperatures for few minutes…

 

[25362]

To rafcu, I've sent the pages. Hope you received them.